Table of standard equivalences 30/57 372 TABLES FOR PART I - PowerPoint PPT Presentation

Table of standard equivalences 30/57 372 TABLES FOR PART I Equivalences for connectives Commutativity: Associativity: val val P ∧ Q = = Q ∧ P , ( P ∧ Q ) ∧ R = = P ∧ ( Q ∧ R ), val val P ∨ Q = = Q ∨ P , ( P ∨ Q ) ∨ R = = P ∨ ( Q ∨ R ), val val P ⇔ Q = = Q ⇔ P ( P ⇔ Q ) ⇔ R = = For the collection of all standard Propositional Logic P ⇔ ( Q ⇔ R ) Idempotence: Double Negation: equivalences, see page 372 of the val val P ∧ P = = P , ¬¬ P = = P Lecture 2 (Chapter 7) val P ∨ P = = P book! Inversion: True/False-elimination: val val ¬ True = = False , P ∧ True = = P , val val ¬ False = = True P ∧ False = = False , val You will have to know them by P ∨ True = = True , val P ∨ False = = P heart (including their names!). Negation: Contradiction: val val ¬ P = = P ⇒ False P ∧ ¬ P = = False Excluded Middle: Start memorising them today! val P ∨ ¬ P = = True September 9, 2016 Distributivity: De Morgan: val val P ∧ ( Q ∨ R ) = = ( P ∧ Q ) ∨ ( P ∧ R ), ¬ ( P ∧ Q ) = = ¬ P ∨ ¬ Q , val val P ∨ ( Q ∧ R ) = = ( P ∨ Q ) ∧ ( P ∨ R ) ¬ ( P ∨ Q ) = = ¬ P ∧ ¬ Q Implication: Contraposition: val val P ⇒ Q = = ¬ P ∨ Q P ⇒ Q = = ¬ Q ⇒ ¬ P Bi-implication: Self-equivalence: val val P ⇔ Q = = ( P ⇒ Q ) ∧ ( Q ⇒ P ) P ⇔ P = = True / department of mathematics and computer science / department of mathematics and computer science val Calculation = is a decent equivalence = 31/57 32/57 Recall the following calculation : Can we conclude ¬ P ⇒ Q val ¬ P ⇒ Q = = P ∨ Q ? val = = { Implication } ¬¬ P ∨ Q Lemma 6.1.1 val = = { Double Negation } val 1. (Reflexivity:) P = = P P ∨ Q val val 2. (Symmetry:) If P = = Q , then Q = = P val val val 3. (Transitivity:) If P = = Q and Q = = R , then P = = R 1. What about applying two standard equivalences in a row? Does it preserve equivalence? 2. First step: not a literal application of Implication. Can we do substitutions? 3. Second step: literal application of Double Negation. Is it safe to apply standard equivalences in a larger context? / department of mathematics and computer science / department of mathematics and computer science

Substitution Substitution 35/57 36/57 Substitution is the replacement of all occurrences of a ‘letter’ Substitution is the replacement of all occurrences of a ‘letter’ by a formula. by a formula. Examples: Examples: 1. If we substitute Q ∧ P for P in the valid equivalence 2. If we substitute ¬ R for Q in the valid equivalence val val P ⇒ Q = = ¬ P ∨ Q , ( Q ∧ P ) ⇒ Q = = ¬ ( Q ∧ P ) ∨ Q , then we get the valid equivalence then we get the valid equivalence val val ( Q ∧ P ) ⇒ Q = = ¬ ( Q ∧ P ) ∨ Q . ( ¬ R ∧ P ) ⇒ ¬ R = = ¬ ( ¬ R ∧ P ) ∨ ¬ R . / department of mathematics and computer science / department of mathematics and computer science Substitution Substitution Rule 37/57 38/57 Substitution is the replacement of all occurrences of a ‘letter’ by a formula. SUBSTITUTION PRESERVES EQUIVALENCE Examples: Important remarks: 3. If we (simultaneously) substitute Q ∧ P for P and ¬ R for Q in the valid equivalence 1. Substitution operates on entire equivalences 2. If you substitute for some letter P in an equivalence, then you val P ⇒ Q = = ¬ P ∨ Q , have to replace all occurrences of P in that equivalence! then we get the valid equivalence val ( Q ∧ P ) ⇒ ¬ R = = ¬ ( Q ∧ P ) ∨ ¬ R . / department of mathematics and computer science / department of mathematics and computer science

Leibniz’s rule Leibniz’s rule 40/57 41/57 Leibniz’s rule is about the replacement of a subformula by Leibniz’s rule is about the replacement of a subformula by an equivalent subformula. an equivalent subformula. Example: Schematically: From the valid equivalence val P ⇒ Q = = ¬ P ∨ Q val P = = Q we can make new valid equivalences by replacing P ⇒ Q in some val · · · P · · · = = · · · Q · · · complex formula by ¬ P ∨ Q , for instance: val ( ¬ P ∧ ( P ⇒ Q )) ∨ R = = ( ¬ P ∧ ( ¬ P ∨ Q )) ∨ R / department of mathematics and computer science / department of mathematics and computer science Proving tautologies—method 1 Proving tautologies—method 1 (example) 42/57 43/57 Prove with a calculation that ¬ ( P ∧ ¬ P ) is a tautology. We have the following calculation: ¬ ( P ∧ ¬ P ) To prove with a calculation that P is a tautology: val = = { De Morgan } ¬ P ∨ ¬¬ P val Give calculation that shows P = = True . val = = { Double Negation } P ∨ ¬ P val = = { Excluded Middle } True So ¬ ( P ∧ ¬ P ) is a tautology. / department of mathematics and computer science / department of mathematics and computer science

Proving tautologies—method 2 Proving tautologies—method 2 (example) 44/57 45/57 Prove with a calculation that ¬ ( Q ⇒ R ) ⇔ ( ¬ R ∧ Q ) is a tautology. Lemma 6.1.3 val First, we establish, with a calculation, that ¬ ( Q ⇒ R ) = = ( ¬ R ∧ Q ) : val If P = = Q , then P ⇔ Q is a tautology, and vice versa. Explanation: ¬ ( Q ⇒ R ) Substituting Q for P and R for Q in val = = { Implication } val To prove with a calculation that P is a tautology: P ⇒ Q = = ¬ P ∨ Q (Implication) ¬ ( ¬ Q ∨ R ) we get, by the substitution rule: val = = { De Morgan } val Q ⇒ R = = ¬ Q ∨ R . ¬¬ Q ∧ ¬ R val Give a calculation that shows P = = Q . (The application of this equivalence in val the calculation involves an application = = { Double negation } of Leibniz.) ¬ R ∧ Q / department of mathematics and computer science / department of mathematics and computer science Proving tautologies—method 2 (example) Proving tautologies—method 2 (example) 45/57 45/57 Prove with a calculation that ¬ ( Q ⇒ R ) ⇔ ( ¬ R ∧ Q ) is a tautology. Prove with a calculation that ¬ ( Q ⇒ R ) ⇔ ( ¬ R ∧ Q ) is a tautology. val val First, we establish, with a calculation, that ¬ ( Q ⇒ R ) = = ( ¬ R ∧ Q ) : First, we establish, with a calculation, that ¬ ( Q ⇒ R ) = = ( ¬ R ∧ Q ) : Explanation: ¬ ( Q ⇒ R ) ¬ ( Q ⇒ R ) Substituting Q for P in Explanation: val val = = { Implication } = = { Implication } val ¬¬ P = = P (Double negation) Substituting ¬ Q for P and R for Q in ¬ ( ¬ Q ∨ R ) ¬ ( ¬ Q ∨ R ) we get, by the substitution rule: val ¬ ( P ∨ Q ) = = ¬ P ∧ ¬ Q val ¬¬ Q = = Q . val val = = { De Morgan } = = { De Morgan } (De Morgan) we get, by the substitution rule: (The application of this equivalence in ¬¬ Q ∧ ¬ R ¬¬ Q ∧ ¬ R val the calculation involves an application ¬ ( ¬ Q ∨ R ) = = ¬¬ Q ∧ ¬ R . val val = = { Double negation } = = { Double negation } of Leibniz, and is followed by an ¬ R ∧ Q ¬ R ∧ Q application of Commutativity.) / department of mathematics and computer science / department of mathematics and computer science

Proving tautologies—method 2 (example) Logical Consequence 45/57 46/57 Recall: Prove with a calculation that ¬ ( Q ⇒ R ) ⇔ ( ¬ R ∧ Q ) is a tautology. � (a) whenever P is 1 , then also Q is 1 val P = = Q means val First, we establish, with a calculation, that ¬ ( Q ⇒ R ) = = ( ¬ R ∧ Q ) : (b) whenever Q is 1 , then also P is 1 ¬ ( Q ⇒ R ) Define: val = = { Implication } ¬ ( ¬ Q ∨ R ) val � P | = = Q (a) whenever P is 1 , then also Q is 1 means val = = { De Morgan } val Pronounce P | = = Q as “ P is stronger than Q .” ¬¬ Q ∧ ¬ R val = = { Double negation } ¬ R ∧ Q P Q 1 1 val P | = = Q : 1 s are carried over from P to Q . val From ¬ ( Q ⇒ R ) = = ¬ R ∧ Q it follows (by Lemma 6.1.3) that 0 1 / 0 ¬ ( Q ⇒ R ) ⇔ ( ¬ R ∧ Q ) is a tautology. / department of mathematics and computer science / department of mathematics and computer science Logical Consequence (example 1) Logical Consequence (example 2) 47/57 48/57 � val � � � val | = = ? = | = ? ¬ P P ⇒ Q P ⇒ Q P ∨ Q val val = = | ? = = | ? P Q ¬ P P ⇒ Q P Q P ⇒ Q P ∨ Q 0 0 1 1 0 0 1 0 extra true 0 1 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 0 1 1 1 1 1 val So P ⇒ Q and P ∨ Q are incomparable. So ¬ P is stronger than P ⇒ Q (i.e., ¬ P | = = P ⇒ Q ). / department of mathematics and computer science / department of mathematics and computer science

val val Standard weakenings Basic properties of | = , | = = = 49/57 50/57 ∧ - ∨ -weakening: P Q P ∧ Q P P ∨ Q val 0 0 0 0 0 P ∧ Q | = = P 0 1 0 0 1 val P | = = P ∨ Q 1 0 0 1 1 Lemma 7.3.1 1 1 1 1 1 Also: val (1a) P | = = P . val val val P ∧ Q | = = Q (2) If P | = = Q , then Q = = | P , and vice versa. and val val val val (3) If P | = = Q and Q | = = R , then P | = = R . Q | = = P ∨ Q . Extremes: val False is strongest of all False | = = P True is weakest of all val P = | = True / department of mathematics and computer science / department of mathematics and computer science val val val val Basic properties of | | Basic properties of | | = , = , = = = = = = 51/57 52/57 Lemma 7.3.2 val val val P = = Q if, and only if, P | = = Q and P = = | Q . Lemma 7.3.4 val P | = = Q if, and only if, P ⇒ Q is a tautology. val val So, if you need to prove P | = = Q or P = = | Q by a calculation, val val val then it is enough to prove P = = Q . But P | = = Q (or P = = | Q ) val alone is not enough to conclude P = = Q ! / department of mathematics and computer science / department of mathematics and computer science